ID CODE
11224 KYFAWT
12215 KYFAHS
12215 KYFAHS
12220 KYFAWT
12501 JHRWSXD
12220 KYFAHS
12210 KYFAHS
12501 KYFAWT
11224 KYFAWT
12220 KYFAWT
写段sql,求每个ID对应不同code的个数
数据库初学者,各位帮帮我啊
------解决方案--------------------
- SQL code
insert into table1(id,code) values('11224','KYFAWT');insert into table1(id,code) values('12215','KYFAHS');insert into table1(id,code) values('12215','KYFAHS');insert into table1(id,code) values('12220','KYFAWT');insert into table1(id,code) values('12501','JHRWSXD');insert into table1(id,code) values('12220','KYFAHS');insert into table1(id,code) values('12210','KYFAHS');insert into table1(id,code) values('12501','KYFAWT');insert into table1(id,code) values('12501','KYFAWT');insert into table1(id,code) values('11224','KYFAWT');insert into table1(id,code) values('12220','KYFAWT');commit;select distinct id,count(distinct code) over (partition by id) from table1 order by id;--结果1 11224 12 12210 13 12215 14 12220 25 12501 2
------解决方案--------------------
- SQL code
SELECT ID, SUM(CASE CODE WHEN 'KYFAWT' THEN 1 ELSE 0 END) AS KYFAWT, SUM(CASE CODE WHEN 'KYFAHS' THEN 1 ELSE 0 END) AS KYFAHS, SUM(CASE CODE WHEN 'JHRWSXD' THEN 1 ELSE 0 END) AS JHRWSXDFROM TABLE1GROUP BY IDORDER BY ID;
------解决方案--------------------
1、如果code确定可以
select id,
sum(decode(code, 'KYFAWT', total, 0)) as code_A,
sum(decode(code, 'KYFAHS', total, 0)) as code_B
...
from (SELECT ID, code, COUNT(1) as total FROM YOUR_TABLE GROUP BY ID, code);
2、如果不确定 采用动态SQL了